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Question

If any tangent plane to the sphere (xa)2+(yb)2+(zc)2=r2 makes intercepts a,b,c with the coordinate axes at A,B,C and P be the center of the sphere. Then

A
1a2+1b2+1c2=4r2
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B
1a2+1b2+1c2=1r2
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C
Volume of the tetrahedron PABC=abc3
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D
Volume of the tetrahedron PABC=abc6
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Solution

The correct options are
A 1a2+1b2+1c2=4r2
C Volume of the tetrahedron PABC=abc3
Equation of the plane is xa+yb+zc=1
Distance of the plane from the center of the sphere is
21a2+1b2+1c2=r1a2+1b2+1c2=4r2
Now in triangle ABC
Distance of C from AB
AB:xaa=y0b=z00=k
any point on the line (a(k+1),bk,0)
Solving the point with the point C
k=a2a2+b2
So perpendicular distance will be
=a2b2a2+b2+c2=a2b2+a2c2+b2c2a2+b2=2abcr(a2+b2)
AB=a2+b2
ar.ABC=122abcr(a2+b2)a2+b2=abcr
perpendicular distance between plane ABC and P
=r
volume =r3abcr=abc3

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